Upload
others
View
0
Download
0
Embed Size (px)
Citation preview
Chapter eleven:p484
Properties of solutions
Contentsp484
11.1 Solution Compositionp485
Mass percent =
Mass fraction of component A = χA =
Molality =solventofkilogram
soluteofmoles
AB
A
nnn
%100)solutionofmasssoluteofmass
(
Table 11.1 Various Types of Solutions
SolidGasSolidHydrogen in platinumLiquidSolidLiquidSeawater, sugar solutionLiquidGasLiquidCarbonated water (soda)SolidSolidSolidBrassLiquidLiquidLiquidVodka in water, antifreezeGasGasGasAir, natural gas
State ofSolvent
State ofSolute
State ofSolutionExample
p485
1. Molarity (M) =
2. Mass (weight) percent =
3. Mole fraction (χA) =
4. Molality (m) =
moles of soluteliters of solution
mass of solutemass of solution
100%
molestotal moles in solution
A
moles of solutekilograms of solvent
Ex 11.1 Various Methods for DescribingSolution Composition
A solution is prepared by mixing 1.00 g ethanol(C2H5OH) with 100.0 g water to give a final volume of101 mL. Calculate the molarity, mass percent, molefraction, and molality of ethanol in this solution.
p486
p487
For oxidation-reduction reaction
Ex 11.2 Calculating various methods of solutioncomposition from the molarity
The electrolyte in automobile lead storage batteries is a
3.75 M sulfuric acid Solution that has a density of
1.230 . Calculate the mass percent, molality, and
normality of the sulfuric acid.
mlg/
Solution:
p487
p488
Since each sulfuric acid ,molecule can furnish twoprotons. 1 mol H2S04 represents 2 equivalent. Thus asolution with 3.75 mol H2SO4 per liter, its normality is7.50 N.
11-2 The Energies of Solution FormationP488
1 Separating the solute into its individual components (expanding the
solute).
2 Overcoming intermolecular forces in the solvent to make room for the
solute(expanding the solvent).
3 Allowing the solute and solvent to interact to form the solution.
321soln HHHH
The Steps in the Dissolving Process
Figure 11.1
p490
Table 11.3 The Energy Terms for VariousTypes of Solutes and Solvents
Nosolutionforms
Large,positive
SmallSmallLargeNonpolar solvent,polar solute
Solutionforms
SmallSmallSmallSmallNonpolar solvent,nonpolar solute
Nosolutionforms
Large,positive
SmallLargeSmallPolar solvent,nonpolar solute
Solutionforms
SmallLarge,negative
LargeLargePolar solvent,polar solute
OutcomeHsolnH3H2H1
p492
Ex 11.3 Differentiating Solvent Properties
Decide whether liquid hexane (C6H14) or liquid methanol
(CH3OH) is the more appropriate solvent for the
substances grease (C20H42) and potassium iodide (KI).Solution:
p492
Think About It
What does the term “hydrophobic”mean?
Oil is sometimes termed “hydrophobic”. Why? Also,
why is the term “hydrophobic”not really correct?
11-3 Factors Affecting Solubilityp492
1. Structural Effects
2. Pressure Effects
3. Temperature Effects
p493
Figure 11.4
The molecular structure of (a) vitamin A (nonpolar,fat-soluble) and vitamin C (polar, water-soluble). Thecircles in the structural formulas indicate polarbonds. Note that vitamin C far more polar bonds thanvitamin A.
A Gaseous Solute
Figure 11.5
p494
Henry’s law C = k P
Ex 11.4 Calculations Using Henry’s Law
A certain soft drink is bottled so that a bottle at 25℃
contains CO2 gas at a pressure of 5.0 atm over the liquid.
Assuming that the partial pressure of CO2 in the
atmosphere is 4.0 × 10-4 atm, calculate the equilibrium
concentration of CO2 in the soda both before and after
the bottle is opened. The Henry’s law constant for CO2 in
aqueous solution is 3.1 × 10-2 at 25℃.atmLmol /
p495
p495
Figure 11.7The solubility of several gases in water as a function oftemperature at constant pressure of 1 atm of gas above thesolution.
p496
11-4 The Vapor Pressure of Solutions
Figure 11.9
An aqueous solution and pure water in a closedenvironment. (a) Initial stage. (b) After a period oftime, the water is transferred to the solution.
p497
Raoult’s law: P soln = χ solvent P solvent
Figure11.10
The presence of a nonvolatile solute inhibits theescape of solvent molecules from the liquid and solowers the vapor pressure of the solvent.
p498
Figure 11.11
A solution obeying raoult’s law, a plot ofPsoln versus solvent gives a straight line.
χ
Ex 11.5 Calculating the Vapor Pressure of aSolution
Calculate the expected vapor pressure at 25 ℃ for asolution prepared by dissolving 158.0 g of commontable sugar (sucrose, molar mass = 342.3 g/mol) in643.5 cm3 of water. At 25 ℃, the density of water is0.9971 g/cm3 and the vapor pressure is 23.76 torr.
p499
p499
Predict the vapor pressure of a solution prepared by mixing
35.0 g solid Na2SO4 (molar mass = 142 g/mol) with 175 g
water at 25℃. The vapor pressure of pure water at 25 ℃ is
23.76 torr.
Ex 11.6 Calculating the Vapor Pressure of aSolution Containing Ionic Solute
Solution:
p500
p501Nonideal Solutions
00BBAABATOTAL PpPPP
Table 11.4 Summary of the Behavior of VariousTypes of Solutions
Ethanol-hexanePositiveNegativePositive
(endothermic)A A, B B > A B
Acetone-waterNegativePositiveNegative
(exothermic)A A, B B < A B
Benzene-toluene
None(ideal
solution)ZeroZeroA A, B B A B
Example
Deviationfrom
Raoult’sLaw
T forSolution
FormationHsoln
Interactive ForcesBetween Solute (A) and
Solvent (B) Particles
p503
p502
Figure 11.13 Vapor pressure for a solution of two volatile liquids. (a)The behavior predicted for an ideal liquid-liquid solution by Raoult;slaw. (b) A solution for which Ptotal is larger than the value calculatedfrom Raoult’s law. (c) A solution for which Ptotal is smaller than thevalue calculated from Raoult’s law. This solurtion shows a negativedeviation from Raoult’s law.
Ex 11.7 Calculating the Vapor Pressure of aSolution Containing Two Liquids
A solution is prepared by mixing 5.81 g acetone (C3H6O,molar mass = 58.1 g/mol) and 11.9 g chloroform (HCCl3,molar mass = 119.4 g/mol). At 35 ℃, this solution has atotal vapor pressure of 260 torr. Is this an ideal solution?The vapor pressure of pure acetone and pure chloroform at35 ℃ are 345 and 293 torr, respectively.
Solution:
p503
p504
Vapor Pressure Lowering: Liquid/Vapor Equilibrium
Vapor Pressure Lowering: Addition of a Solute
Vapor Pressure Lowering: Solution/Vapor Equilibrium
Colligative Properties
Depend only on the number, not on the identity, of
the solute particles in an ideal solution.
Boiling point elevation
Freezing point depression
Osmotic pressure
11-5 Boiling-Point Elevation andFreezing-Points Depression
p504
Boiling-Point Elevation
Changes in Boiling Point and Freezing Point of Water
Figure 11.14 Phase diagrams for pure water (red lines) and for anaqueous solution containing a nonvolatile solute (blue lines)
p504
Boiling-Point Elevation: Liquid/Vapor Equilibrium
Boiling-Point Elevation: Addition of a Solute
Boiling Point Elevation: Solution/Vapor Equilibrium
Freezing Point Depression: Solid/Liquid Equilibrium
Freezing Point Depression: Solid/Solution Equilibrium
p505
46
Ex 11.8 Calculating the Molar Mass byBoiling Point Elevation
A solution was prepared by dissolving 18.00 gglucose in 150.0 g water. The resulting solution wasfound to have a boiling point 100.34℃. Calculatethe molar mass of glucose. Glucose is a molecularsolid that is present as individual molecules insolution.solution
Thus 18.00 g glucose has0.10 mol, and 1.0 mol has a mass of 180 g.
p506
Freezing-Points Depression
p506
48
Ex 11.9 Freezing-Point DepressionWhat mass of ethylene glycol (C2H6O2, molar mass =62.1 g/mol), the main component of antifreeze, must beadded to 10.0 L water to produce a solution for use in acar’s radiator that freezes at -23.3℃ ? Assume thedensity of water is exactly 1 g/mL.
Solution:
p506
49
Ex 11.10 Determining Molar Mass by Freezing-Point Depression
A chemist is trying to identify a human hormone that
controls metabolism by determining its molar mass. A
sample weighing 0.546 g was dissolved in 15.0 g benzene,
and the freezing-point depression was determined to be
0.240℃. Calculate the molar mass of the hormone.
p507
p507
Freezing Point Depression: Addition of a Solute
11-6 Osmotic Pressure p508
Semipermeable membrane
Osmotic Pressure
Figure 11.6 A tube with a bulb on the end that iscovered by a semipermeable membrane.
p508
Osmosis
p509
Ex 11.11 Determining Molar Mass fromOsmotic Pressure
To determine the molar mass of a certain pertain,1.00 × 10-3 g of it was dissolved in enough water tomake 1.00 mL of solution. The osmotic pressure ofthis solution was found to be 1.12 torr at 25.0 ℃.Calculate the molar mass of the protein.Solution:
p509
Ex 11.12 Isotonic SolutionsWhat concentration of sodium chloride in water is
needed to produce an aqueous solution isotonic with
blood ( = 7.70 atm at 25℃)?
p510
p511
60
Reverse Osmosisp511
11-7 Colligative Properties ofElectrolyte Solutions
p512
The relationship between the moles solutes
dissolved and the moles particles in solution is
usually expressed using the van’t Hoff factor, i:
dissolvedsoluteofmolessolutioninparticlesofmoles
i
p512
Ex 11.13 Osmotic PressureThe observed osmotic pressure for a 0.10 M solution of
Fe(NH4)2(SO4)2 at 25 ℃ is 10.8 atm. Compare the expected
and experimental values for i.
Solution
p513
65
11.8 Colloidsp515